Electron Configuration and Effective Nuclear Charge PDF

Electron Configurations  Learning Outcomes At the end of this topic, students should be able to:  State and apply Aufbau principle, Hund's rule and Pauli exclusion principle in filling of electrons in orbitals of an atom.  Write the electronic configuration of...

Electron Configurations  Learning Outcomes At the end of this topic, students should be able to:  State and apply Aufbau principle, Hund's rule and Pauli exclusion principle in filling of electrons in orbitals of an atom.  Write the electronic configuration of atoms and monoatomic ions. (a) Orbital diagram (b) spdf notation Explain the anomalous electronic configurations of chromium and copper.  Electron Configuration  The electron configuration describes how the electrons are distributed in the various atomic orbitals.  In a ground state hydrogen atom, the electron is found in the 1s orbital. Ground state electron configuration of hydrogen principal quantum number (n = 1) number of electrons in the 1s 1 orbital or subshell Energy 2s 2p 2p 2p angular momentum quantum number (l = 0) The use of an up arrow indicates an 1s electron with ms = + ½  Electron Configuration  If hydrogen’s electron is found in a higher energy orbital, the atom is in an excited state. A possible excited state electron configuration of hydrogen Energy 2s 2p 2p 2p 2s1 An electron is normally in its ground state, the lowest energy state available. After absorbing energy, it may jump from the 1s ground state to a higher energy level, called an excited state. An excited state is an energy level of an atom, ion, or molecule in which an electron is at a higher energy level than its ground state.  Orbital Diagrams  Show the arrangement of electrons in orbitals within an atom.  Use boxes , circle or line to represent orbitals.  One arrow (↑) represents 1 e-.  2 arrows (↑↓) represents 2 e-. Abbreviated Orbital Diagrams Ru [Kr]       4d 5s S [Ne]     3s 3p  Pauli Exclusion Principle  According to the Pauli exclusion principle, no two electrons in an atom can have the same four quantum numbers. The ground state electron configuration of helium 2p 2p 2p 1s2 Energy 2s Quantum number Principal (n) 1 1 1s describes the 1s orbital Angular moment (l) 0 0 Magnetic (ml) 0 0 describes the electrons in the 1s orbital Electron spin (ms) +½ ‒½  Aufbau Principle  The Aufbau principle states that electrons are added to the lowest energy orbitals first before moving to higher energy orbitals. Li has a total of 3 electrons The ground state electron configuration of Li 2p 2p 2p 1s22s1 Energy 2s The third electron must go in the next available orbital with the lowest possible energy. The 1s orbital can only accommodate 2 electrons 1s (Pauli exclusion principle) Increasing Energy Aufbau Principle  Hund’s Rule Hund’s rule states that single electrons must occupy each equal-energy orbital before additional electrons with can occupy the same energy level orbitals. If you have more than one orbital all at the same energy Put one electron into each orbital with parallel spins (↑↑) until all are half filled (All electrons in singly occupied orbitals must have the same spin). After orbitals are half full, follow by pair up electrons. 134  Hund’s Rule  According to Hund’s rule, the most stable arrangement of electrons is the one in which the number of electrons with the same spin is maximized. The ground state electron C has a total of 6 configuration of C electrons 1s22s22p2 2p 2p 2p Energy 2s The 2p orbitals are of equal energy, or degenerate. Put 1 electron in each before pairing (Hund’s rule). 1s  Hund’s Rule  According to Hund’s rule, the most stable arrangement of electrons is the one in which the number of electrons with the same spin is maximized. The ground state electron N has a total of 7 configuration of N electrons 1s22s22p3 2p 2p 2p Energy 2s The 2p orbitals are of equal energy, or degenerate. Put 1 electron in each before pairing (Hund’s rule). 1s  Hund’s Rule  According to Hund’s rule, the most stable arrangement of electrons is the one in which the number of electrons with the same spin is maximized. The ground state electron O has a total of 8 configuration of O electrons 1s22s22p4 2p 2p 2p Energy 2s Once all the 2p orbitals are singly occupied, additional electrons will have to pair with those already in the orbitals. 1s  Ground State Electron Arrangements Electron Configurations Distribution of electrons among orbitals of atom 1. List subshells that contain electrons 2. Indicate their electron population with superscript e.g., N is 1s2 2s2 2p3 Orbital Diagrams Way to represent electrons in orbitals 1. Represent each orbital with circle (or line) 2. Use arrows to indicate spin of each electron e.g., N is 1s 2s 2p  Know from Magnetic Properties Two electrons in same orbital have different spins Spins paired - diamagnetic Sample not attracted to magnetic field Magnetic effects tend to cancel each other Two electrons in different orbital with same spin Spins unpaired - paramagnetic Sample attracted to a magnetic field More unpaired electrons lead to a stronger attraction to a magnet Measure extent of attraction Gives number of unpaired spins 139  Learning Check Give electron configurations and orbital diagrams for Na (Z = 11) and As (Z = 33) and determine if they are para- or diamagnetic 6s 5p 4d 5s 4p 4s 3d 3p 3s Energy 2p 2s Na Z = 11 1s2 2s2 2p6 3s1 paramagnetic As Z = 33 1s22s22p63s23p64s23d104p3 paramagnetic 1s  Your Turn! What is the correct ground state electron configuration for Si (Z = 14)? A. 1s22s22p63s23p6 B. 1s22s22p63s23p4 C. 1s22s22p62d4 D. 1s22s22p63s23p2 E. 1s22s22p63s13p3  The Anomalous Electronic Configurations of Cr and Cu  Cr and Cu have electron configurations which are inconsistent with the Aufbau principle. The anomalous are explained on the basis that a filled or half-filled orbital is more stable. Element Expected (Aufbau Principle) Observed/actual Cr (Z = 24) 1s22s22p63s23p6 4s2 3d4 1s22s22p63s23p6 4s1 3d5 Cu (Z = 29) 1s22s22p63s23p6 3d9 4s2 1s22s22p63s23p6 3d10 4s1  Electronic Configurations A few exceptions to rules Element Expected Experimental Cr [Ar] 3d 44s 2 [Ar] 3d 54s 1 Cu [Ar] 3d 94s 2 [Ar] 3d 104s 1 Ag [Kr] 4d 95s 2 [Kr] 4d 105s 1 Au [Xe] 5d 96s 2 [Xe] 5d 106s 1  Exactly filled and exactly half-filled subshells have extra stability  Promote one electron into ns orbital to gain this extra stability Jespersen/Hyslop, Chemistry: The Molecular Nature of Matter, 7E, Copyright 2015 John Wiley &Sons, Inc. There are several notable exceptions to the order of electron filling for some of the transition metals.  Chromium (Z = 24) is [Ar]4s13d5 and not [Ar]4s23d4 as expected.  Copper (Z = 29) is [Ar]4s13d10 and not [Ar]4s23d9 as expected. The reason for these anomalies is the slightly greater stability of d subshells that are either half-filled (d5) or completely filled (d10). Cr [Ar] 4s 3d 3d 3d 3d 3d Greater stability with half-filled 3d subshell  Abbreviated Electron Configurations - Noble Gas Notation  Noble gas notation uses noble gas symbols in brackets to shorten inner electron configurations of other elements.  The noble gas must have a lower atomic number than the atom or ion that the electron configuration is being written for.  [noble gas of previous row] and electrons filled in next row  Represents core + outer shell electrons  Use to emphasize that only outer shell electrons react e.g., Ba = [Xe] 6s2 Ru = [Kr] 4d6 5s2 S = [Ne] 3s2 3p4  Noble Gas Core Notation for Mn  Find last noble gas that is filled before Mn  Next fill sublevels that follow [Ar] 4s 2 3d 5 n= 1 1 “ns” orbital being filled 2 H He n= 2 3 4 “np” orbital being filled 5 6 7 8 9 10 Li Be “(n – 1)d” orbital being filled B C N O F Ne n= 3 11 12 “( n – 2)f” orbital being filled 13 14 15 16 17 18 Na Mg Al Si P S Cl Ar n= 4 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr n= 5 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe n= 6 55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn n= 7 87 88 89 104 105 106 107 108 109 110 111 Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Rg 58 59 60 61 62 63 64 65 66 67 68 69 70 71 Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu 90 91 92 93 94 95 96 97 98 99 100 101 102 103 Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr  Noble Gas Notation  The electron configurations of all elements except hydrogen and helium can be represented using a noble gas core.  The electron configuration of potassium (Z = 19) is 1s22s22p63s23p64s1.  Because 1s22s22p63s23p6 is the electron configuration of argon, we can simplify potassium’s to [Ar]4s1. The ground state electron configuration of K: 1s22s22p63s23p64s1 [Ar] [Ar]4s1  Look at Group 2 Z Electron Configuration Abbrev Be 4 1s22s2 [He] 2s2 Mg 12 1s22s22p632 [Ne] 3s2 Ca 20 1s22s22p63s23p64s2 [Ar] 4s2 Sr 38 1s22s22p63s23p63d104s24p65s2 [Kr] 5s2 Ba 56 1s22s22p63s23p63d104s24p64d105s25p66s2 [Xe] 6s2 Ra 88 1s22s22p63s23p63d104s24p64d104f145s25p65d106s26p67s2 [Rn] 7s2  All have ns2 outer shell electrons  Only difference is value of n  Your Turn! The ground state electron configuration for Ca is: a) [Ar] 3s1 n= 1 1 2 b) 1s2 2s2 2p6 3s2 3p5 4s2 n= 2 H3 4 5 6 7 8 9 He 10 Li Be B C N O F Ne c) [Ar] 4s 2 n= 3 11 12 13 14 15 16 17 18 Na Mg Al Si P S Cl Ar d) [Kr] 4s1 n= 4 19 20 21 22 23 24 25 26 27 K Ca Sc Ti V Cr Mn Fe Co 28 29 30 31 32 33 Ni Cu Zn Ga Ge As Se Br Kr 34 35 36 n= 5 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 e) [Kr] 4s2 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe n= 6 55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn n= 7 87 88 89 104 105 106 107 108 109 110 111 Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Rg 58 59 60 61 62 63 64 65 66 67 68 69 70 71 Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu 90 91 92 93 94 95 96 97 98 99 100 101 102 103 Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr  Pseudo noble gas electronic configuration Noble gas electronic configuration: ends in ns2np6 Pseudo noble gas electronic configuration: ends in ns2np6nd10  Applies to transition metal cations, such as Ag+, Zn2+ Ag: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s1  Your Turn! Which of the following elements will commonly ionize to attain pseudo noble gas electronic configuration? Cu, Cr, V, Cd, Mo, Sn Solution Cu (Cu+) Cu: 1s2 2s2 2p6 3s2 3p6 3d10 4s1 Cd (Cd2+) Cd: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 Sn (Sn4+) Sn: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p2  Your Turn! The ground state electron configuration for Ca is: a) [Ar] 3s1 b) 1s2 2s2 2p6 3s2 3p5 4s2 c) [Ar] 4s2 d) [Kr] 4s1 e) [Kr] 4s2  Your Turn! Which of the following choices is the correct orbital diagram for a cobalt atom? 4s 3d A. [Ar] ↑↓ ↑↓ ↑↓ ↑↓ ↑ B. [Ar] ↑ ↑↓ ↑↓ ↑↓ ↑↓ C. [Ar] ↑ ↑↓ ↑↓ ↑↓ ↑ ↑ D. [Ar] ↑↓ ↑↓ ↑↓ ↑↓ ↑ E. [Ar] ↑↓ ↑↓ ↑↓ ↑ ↑ ↑  Your Turn! Which one of the following represents an excited state electron configuration of C (Z = 6)? a) 1s2 2s2 2p3 b) 1s2 2s2 2p2 c) 1s2 2s2 2p1 3s1 d) 1s2 2s1 2p3  Your Turn! If the electronic configuration of nitrogen (Z = 7) is written as 1s2 2s2 2px2 2py1, it would violate a) Aufbau principle b) Pauli’s exclusion principle c) Hund’s rule of maximum multiplicity d) none of these  Your Turn! If the Aufbau principle had not been followed, the last electron of Ca (Z = 20) would have been placed in the: a) s–block b) p–block c) d–block d) f–block  Your Turn! The px, py and pz orbitals are called degenerate orbitals as they have a) equal energy b) same orientation in space c) same size d) none of these  Your Turn! Which is the ground-state electronic configuration of Na+? a) 1s2 2s2 2p8 b) 1s2 2s2 2p6 3s1 Na = 1s2 2s2 2p6 3s1 c) 1s2 2s2 2p6 3s2 Na+ = 1s2 2s2 2p6 d) 1s2 2s2 2p6  Your Turn! Which of the following does not have the ground-state configuration 1s2 2s2 2p6 a) Ne b) Na+ c) Cl– d) F–  Your Turn! Which of the following atoms or ions has the electronic configuration 1s2 2s2 2p6 3s2 3p6? a) Ne b) K c) Cl– d) Mg2+  Your Turn! In its ground state, an H– ion has: a) two spin-paired electrons b) two electrons with parallel spins c) two electrons in different orbitals d) one electron  Your Turn! According to Aufbau’s principle, which of the three 4d, 5p and 5s will be filled with electrons first a) 4d b) 5p c) 5s d) 4d and 5s will be filled simultaneously  Your Turn! The electronic configuration of an element is 1s2 2s2 2p6 3s2 3p6 3d5 4s1. This represents its a) Excited state b) Ground state c) Cationic form d) Anionic form  Your Turn! The orbital diagram corresponding to the ground state electron configuration for nitrogen is: A. 1s 2s 2p B. 1s 2s 2p C. 1s 2s 2p D. E. 1s 2s 2p 1s 2s 2p  Electron shielding and effective nuclear charge  Learning Outcomes At the end of this topic, students should be able to:  Describe how to calculate the shielding constant using Slater's rules  Identify a periodic trend for effective nuclear charge  Calculate effective nuclear charge  Electron Shielding and Effective nuclear charge  Those electrons in the outmost or valence shell are especially important because they are the ones that can engage in the sharing and exchange that is responsible for chemical reactions;  how tightly they are bound to the atom determines much of the chemistry of the element.  The degree of binding is the result of two opposing forces:  the attraction between the electron and the nucleus.  the repulsions between the electron in question and all the other electrons in the atom. Inner/Core electrons Z Valence electrons Force of repulsion Jespersen/Hyslop, Chemistry: The Molecular Nature of Matter, 7E, Copyright 2015 John Wiley &Sons, Inc. Force of attraction A B  Electron Shielding  Electrons in atoms are attracted to the nucleus, and at the same time repelled by other electrons in the atom.  Electron shielding occurs when the inner shielding electrons shield the outer valence electrons from the full attraction of the nucleus.  The valence electron(s) within an atom require less energy to remove than the inner electrons. Jespersen/Hyslop, Chemistry: The Molecular Nature of Matter, 7E, Copyright 2015 John Wiley &Sons, Inc.  Electron Shielding  Electron shielding remains constant across a period (left to right) Na 1s2 2s2 2p6 3s1 Mg 1s2 2s2 2p6 3s2 Al 1s2 2s2 2p6 3s2 3p1 Si 1s2 2s2 2p6 3s2 3p2 P 1s2 2s2 2p6 3s2 3p3 S 1s2 2s2 2p6 3s2 3p4 Cl 1s2 2s2 2p6 3s2 3p5 Ar 1s2 2s2 2p6 3s2 3p6 Jespersen/Hyslop, Chemistry: The Molecular Nature of Matter, 7E, Copyright 2015 John Wiley &Sons, Inc.  Electron Shielding  Electron shielding Li 1s2 2s1 increases down a group Na 1s2 2s2 2p6 3s1 K 1s2 2s2 2p6 3s2 3p6 4s1 Jespersen/Hyslop, Chemistry: The Molecular Nature of Matter, 7E, Copyright 2015 John Wiley &Sons, Inc. Rb 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s1  Effective nuclear charge  Three protons in nucleus  Two core electrons in close (1s)  Net positive charge felt by outer electron Approximately one proton Effective Nuclear Charge (Zeff) Net positive charge outer electron feels Core electrons shield valence electrons from full nuclear charge Zeff = Z – S where Z is the atomic number and S is the number of shielding electrons Note that this equation only gives an approximate value for the effective nuclear charge of an atom. Jespersen/Hyslop, Chemistry: The Molecular Nature of Matter, 7E, Copyright 2015 John Wiley &Sons, Inc.  Effective nuclear charge  Effective Nuclear Charge (Zeff) increases across a period (left to right) Na 1s2 2s2 2p6 3s1 Zeff (Na) = 11 – 10 = +1 Mg 1s2 2s2 2p6 3s2 Zeff (Mg) = 12 – 10 = +2 Al 1s2 2s2 2p6 3s2 3p1 Zeff (Al) = 13 – 10 = +3 Si 1s2 2s2 2p6 3s2 3p2 Zeff (Si) = 14 – 10 = +4 P 1s2 2s2 2p6 3s2 3p3 Zeff (P) = 15 – 10 = +5 S 1s2 2s2 2p6 3s2 3p4 Zeff (S) = 16 – 10 = +6 Cl 1s2 2s2 2p6 3s2 3p5 Zeff (Cl) = 17 – 10 = +7 Ar 1s2 2s2 2p6 3s2 3p6 Zeff (Ar) = 18 – 10 = +8 Jespersen/Hyslop, Chemistry: The Molecular Nature of Matter, 7E, Copyright 2015 John Wiley &Sons, Inc.  Effective nuclear charge  Effective Nuclear Charge (Zeff) remains the same down a group. Li 1s2 2s1 Zeff (Li) = 3 – 2 = +1 Na 1s2 2s2 2p6 3s1 Zeff (Na) = 11 – 10 = +1 K 1s2 2s2 2p6 3s2 3p6 4s1 Zeff (K) = 19 – 18 = +1 Rb 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s1 Zeff (Rb) = 37 – 36 = +1 Jespersen/Hyslop, Chemistry: The Molecular Nature of Matter, 7E, Copyright 2015 John Wiley &Sons, Inc. Worked Example What is the effective nuclear charge for an electron in the n = 2 shell in an atom of Fluorine? Solution Z=9 Zeff = Z – S 9 p+ =9–2 = +7 Your Turn! What is the effective nuclear charge felt by an electron in the n = 3 shell of sulphur Solution Z = 16 Zeff = Z – S = 16 – 10 = +6 16 p+ Your Turn! What is the effective nuclear charge felt by an electron in the n = 2 shell of chlorine Solution Z = 17 Zeff = Z – S n=3 n=2 = 17 – 2 n=1 = +15 17 p+  Your Turn! What value is the closest estimate of Zeff for a valence electron of the calcium atom? A. 1 B. 2 Zeff = 20 protons – 18 core electrons = 2 C. 6 D. 20 E. 40  Effective nuclear charge and Slater’s rules  Slater's rules allow you to estimate the effective nuclear charge (Zeff) from the real number of protons in the nucleus and the effective shielding of electrons in each orbital.  Effective nuclear charges, Zeff, experienced by electrons in different atomic orbitals may be estimated using Slater’s rules: Zeff = Z - σ where Z = nuclear charge (atomic number), Zeff = effective nuclear charge, σ = screening (or shielding) constant.  Values of σ may be estimated as follows:  Step 1: Write the electron configuration of the atom in the following form: (1s) (2s, 2p) (3s, 3p) (3d) (4s, 4p) (4d) (4f) (5s, 5p) …  Effective nuclear charge and Slater’s rules Step 2: Identify the electron of interest, and ignore all electrons in higher groups (to the right in the list from Step 1). These do not shield electrons in lower groups. Step 3: Slater's Rules is now broken into two cases: the shielding experienced by an s- or p- electron,  electrons within same group shield by 0.35, except the 1s which shield by 0.30  electrons within the n-1 group shield by 0.85  electrons within the n-2 or lower groups shield by 1.00 the shielding experienced by nd or nf valence electrons  electrons within same group shield by 0.35  electrons within the lower groups shield by 1.00  Effective nuclear charge and Slater’s rules Strategy s and p per electron electrons 0.35 in an orbital 1. Determine the electron configuration and identify the electron of interest. 0.85 2. Use the appropriate Slater Rule to calculate 1.00 the shielding constant for the electron. 1.00 3. Determine the effective nuclear constant. n-3 n-2 n-1 n 1.00 Rules to determine shielding based on Slater's 1.00 1.00 values: 0.35 d and f electrons A general equation: σ = 1.00(#e-)n-2+0.85(#e-)n-1+0.35(#e-)n0 per electron in an orbital Worked example What is the shielding constant experienced by a 2p electron in the nitrogen atom? Solution A. Determine the electron configuration of nitrogen, then write it in the appropriate form. N: 1s2 2s2 2p3 B. Use the appropriate Slater Rules to calculate the shielding constant for the electron. N: (1s2)(2s2,2p3) (i) There are 4 other electrons in the same ns, np group @ 0.35: (2s2,2p3) (ii) There are 2 electrons in the n-1 groups @ 0.85: (1s2) σ[2p] = 0.85(2) + 0.35(4) = 3.10 1s electrons 2s and 2p electrons Worked example Calculate the effective nuclear of the valence electron for oxygen: O, Z = 8 Solution Electron configuration: 1s2 2s2 2p4 a) (1s2) (2s2 2p4) b) σ = (2 x 0.85) + (5 x 0.35) = 3.45 1s 2s, 2p Zeff = Z – σ Zeff = 8 – 3.45 = 4.55 Worked example What is the effective nuclear charge experienced by a valence p-electron in boron? Solution Step A B: 1s2 2s2 2p1. The valence p- electron in boron resides in the 2p subshell. B: (1s2)(2s2,2p1) σ[2p] = 1.00(0) + 0.85(2) + 0.35(2) = 2.40 Step C Z = 5 Zeff = 5 – 2.40 = 2.60 Worked example Calculate the effective nuclear charge experience by 4s electron in potassium atom? Solution K = 19 1s2 (2s2 2p6) (3s2 3p6) 4s1 n–3 n–2 n–1 n σ = (0.35 x 0) + (0.85 x 8) + (1 x 10) = 6.8 + 10 = 16.8 Zeff = Z – σ = 19 – 16.8 = 2.2 Worked example Calculate Zeff for (i) the last electron in Cl atom (Z = 17) and (ii) 1s electron in N atom (Z = 7). Solution (i) for Cl (Z = 17), 17Cl = 1s2 2s22p6 3s23p5 2 8 7 Last shell contains 7 electrons, hence remaining electrons in this shell for the last electron = 7 – 1 = 6 σ for the last electron of Cl atom = (0.35 x 6) + (0.85 x 8)+ (1.0 x 2) = 10.90 Zeff for the last electron of Cl atom = 17 – 10.90 = 6.10 (ii) For N (Z = 7), 7N = 1s2 2s22p3 2 5 There will be no contribution of valence shell electrons towards σ of 1s1 electron. σ of 1s1 electron of N atom = (0 x 5) + (0.30 x 1) = 0.30 and Zeff for this electron = Z – σ = 6.70 Example with two electrons for nickel: Ni, Z = 28 Electron configuration: 1s2 2s2 2p6 3s2 3p6 3d8 4s2 (1s2) (2s2 2p6) (3s2 3p6) (3d8) (4s2) Rearrange them into the same subshell For a 3d electron: σ = (18 x 1.00) + (7 x 0.35) = 20.45 1s,2s,2p,3s,3p 3d Zeff = Z – σ Zeff = 28 – 20.45 = 7.55 For a 4s electron: σ = (10 x 1.00) + (16 x 0.85) + (1 x 0.35) = 23.95 1s,2s,2p 3s,3p,3d 4s Zeff = Z – σ Zeff = 28 – 23.95 = 4.05  Your Turn! What is the shielding constant experienced by a valence p-electron in the bromine atom (Z = 35)? Solution A. Determine the electron configuration of bromine, then write it in the appropriate form. B. Use the appropriate Slater Rules to calculate the shielding constant for the electron. Step A Br: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5 Br: (1s2)(2s2,2p6)(3s2,3p6)(3d10)(4s2,4p5) There are 6 other electrons in the same ns, np group There are 18 electrons in the n-1 groups: (3s, 3p and 3d) There are 10 electrons in the n-2 and lower groups: (1s, 2s and 2p) σ = 6(0.35) + 18(0.85) + 10(1.00) = 27.4  Your Turn! What is the shielding constant experienced by a 3d electron in the bromine atom (Z = 35)? Solution Br: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5 Br: (1s2)(2s2,2p6)(3s2,3p6)(3d10)(4s2,4p5) There are 9 other electrons in the same ns, np group @ 0.35: (3s2,3p6) There are 18 other lower group elements electrons in the same nd, nf group @ 1: (1s2)(2s2,2p6) Ignore the group to the right of the 3d electrons; these do not contribute to the shielding constant @ 0: (4s2,4p5) σ[3d] = 1.00(18) + 0.35(9) = 21.15 Worked example Calculate σ and Zeff values for 4s electron in (i) Cu (Z = 29) and (ii) Mn (Z = 25) atom. Solution (i) for cu (Z = 29), 29Cu = 1s22s22p6 3s23p63d10 4s1 10 18 1 σ for 4s electron in Cu atom = (0.35 x 0) + (0.85 x 18) + (1.0 x 10) = 25.30 Zeff = Z – σ = 29 – 25.30 = 3.70 (ii) For Mn (Z = 25), 25Mn = 1s22s22p6 3s23p63d5 4s2 10 13 2 σ for 4s electron in Mn-atom = (0.35 x 1) + (0.85 x 13) + (1.0 x 10) = 21.40 Zeff = Z – σ = 25 – 21.40 = 3.60 Worked example Calculate σ and Zeff for 3d electron in (i) Cr (Z = 24) and (ii) Ni (Z = 28). Solution (i) (i) for Cr (Z = 24), 24Cr = 1s22s22p63s23p6 3d5 4s1 18 5 1 Here 4s1 electron does not contribute towards the value of σ. σ for a 3d-electron in Cr atom = (0.35 x 4) + (1.0 x 18) = 19.40 (remaining electrons in 3d orbitals = 5 – 1 = 4) Zeff experienced by a 3d electron in Cr atom = Z – σ = 24 – 19.40 = 4.60 (ii) For Ni (Z = 28), 28Ni = 1s22s22p63s23p6 3d8 4s2 18 8 2 4s2 electrons do not contribute to the value of σ. Remaining electrons in 3d orbitals = 8 – 1 = 7. σ for a 3d-elctron in Ni atom = (0.35 x 7) + (1.0 x 18) = 20.45 Zeff for a 3d-electron in this atom = Z – σ = 28 – 20.45 = 7.55  Your Turn! Effective nuclear charge on last electron of Fe2+ a) 7.5 b) 7.45 c) 6.25 d) 19.75